Thread: analysis-bounded monotone sequence problem

Prove that the sequence
lim n-->infinity 1/n(1 + 1/2 + ...+ 1/n) = 0

is
(i) monotone
(ii) bounded
(iii) find its limits

How can i proceed to prove that it is monotone and bounded???
Help me please!
Thank You!

Is 1 + 1/2 + ... + 1/n in the denominator?

If it is in the numerator, then this is divergent, since it is greater than 1/n, which is divergent.

(EDIT: nevermind this. I thought it was a sum!)

If 1 + 1/2 + ... + 1/n is in the denominator, then the denominator itself is increasing (at each step the sum gets bigger, and it gets multiplier by a greater number). This implies that the sequence is decreasing. It is bounded below since all its terms are positive. And it will converge to 0, since it is smaller than 1/n, which converges to 0.

Thank You.
It is 1/n * (1 + ... )

How about the monotone?

Thank You.
It is 1/n * (1 + ... )

How about the monotone?

an = (1/n) * (1 + 1/2 + ... + 1/n) < (1 + 1/4 + ... +1/n^2)

this last sequence is convergent since it is the series with term 1/k^2. This proves that your sequence is bounded.

a_n = (1/n) * (1 + 1/2 + ... + 1/n)

a_(n+1) = 1/(n+1) * (n*a_n + 1/(n+1))

(n+1) * ( a_(n+1) - a_n ) = 1/(n+1) - a_n = 1/(n+1) - (1/n) * (1 + 1/2 + ... +1/n) < 1/(n+1) - 1/n < 0

This proves that the sequence is monotone.

The last expression tends to zero, since it is the product of a convergence sequence (the series) and an infinitesimal one. This proves that a_n tends to zero

This is a subtopic of the problem of sequence of means.
Suppose that is a sequence define .
Theorem: If monotonic then is monotonic.
Theorem: If then .